Solve (5x-1)^2=16

$$( 5 x - 1 ) ^ { 2 } = 16$$

Solve for x

$$x=1$$

$$x=-\frac{3}{5}=-0.6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=16$$

Subtract 16 from both sides.

$$25x^{2}-10x+1-16=0$$

Subtract 16 from 1 to get -15.

$$25x^{2}-10x-15=0$$

Divide both sides by 5.

$$5x^{2}-2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=5\left(-3\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 5x^{2}-2x-3 as \left(5x^{2}-5x\right)+\left(3x-3\right).

$$\left(5x^{2}-5x\right)+\left(3x-3\right)$$

Factor out 5x in the first and 3 in the second group.

$$5x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(5x+3\right)$$

To find equation solutions, solve x-1=0 and 5x+3=0.

$$x=1 x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=16$$

Subtract 16 from both sides.

$$25x^{2}-10x+1-16=0$$

Subtract 16 from 1 to get -15.

$$25x^{2}-10x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -10 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25\left(-15\right)}}{2\times 25}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25\left(-15\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-10\right)±\sqrt{100-100\left(-15\right)}}{2\times 25}$$

Multiply -100 times -15.

$$x=\frac{-\left(-10\right)±\sqrt{100+1500}}{2\times 25}$$

Add 100 to 1500.

$$x=\frac{-\left(-10\right)±\sqrt{1600}}{2\times 25}$$

Take the square root of 1600.

$$x=\frac{-\left(-10\right)±40}{2\times 25}$$

The opposite of -10 is 10.

$$x=\frac{10±40}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{10±40}{50}$$

Now solve the equation x=\frac{10±40}{50} when ± is plus. Add 10 to 40.

$$x=\frac{50}{50}$$

Divide 50 by 50.

$$x=1$$

Now solve the equation x=\frac{10±40}{50} when ± is minus. Subtract 40 from 10.

$$x=-\frac{30}{50}$$

Reduce the fraction \frac{-30}{50} to lowest terms by extracting and canceling out 10.

$$x=-\frac{3}{5}$$

The equation is now solved.

$$x=1 x=-\frac{3}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1=16$$

Subtract 1 from both sides.

$$25x^{2}-10x=16-1$$

Subtract 1 from 16 to get 15.

$$25x^{2}-10x=15$$

Divide both sides by 25.

$$\frac{25x^{2}-10x}{25}=\frac{15}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{10}{25}\right)x=\frac{15}{25}$$

Reduce the fraction \frac{-10}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{2}{5}x=\frac{15}{25}$$

Reduce the fraction \frac{15}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{2}{5}x=\frac{3}{5}$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=\frac{3}{5}+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{3}{5}+\frac{1}{25}$$

Add \frac{3}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{16}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{4}{5} x-\frac{1}{5}=-\frac{4}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=1 x=-\frac{3}{5}$$